Answer:
Density = 2.09×10⁻³ g/mL
Explanation:
We solve this problem from the Ideal Gases Law.
P . V = n . R . T
We know that in STP conditions 1 mol of any gas occupies 22.4 L. So we can compare both situations.
22.4 L = 22400 mL
(P₁ . V₁) / T₁ = (P₂ . V₂) / T₂
We make some conversions:
97°C + 273 = 370 K
755 mmHg . 1 atm / 760mmHg = 0.993 atm
We replace data: (1 atm . 22400 mL) / 273 K = (0.993 atm . V₂) / 370 K
((1 atm . 22400 mL) / 273 K) . 370K = 0.993 atm . V₂
V₂ = 30359 atm . mL / 0.993 atm → 30573 mL
Under those conditions (97 °C and 755 mm Hg) 1 mol of gas is contained in 30573 mL.
We determine the amount of gas: 1 mol of SO₂ weighs 64.06g
Density = m/V → 64.06 g / 30573 mL = 2.09×10⁻³ g/mL
